I would do this through another route;I agree with Devron with 0.302g Na2SO4 used and the percentage comes out the same but uses different approaches.. Take you pick.
1.281 = g Na2SO4 initially. Some of that reacts to form Na2SO4.10H2O and some of it remains as Na2SO4. We need to find the two parts; i.e., the amount remaining intact and the amount used to make Na2SO4.1OH2O.

(Note: mm stands for molar mass.)
How much will the Na2SO4.10H2O weigh? That will be 0.383 x (mm Na2SO4.10H2O/mm10H2O) = 0.383 x (322.196/10*18.015) = about 0.685g Na2SO4.10H2O.